1. Key Concepts and Formulas

• Combinations ($^n C_r$): Used to calculate the number of ways to choose $r$ items from a set of $n$ items when the order of selection does not matter. The formula is $^n C_r = \frac{n!}{r!(n-r)!}$.
• Total Probability Theorem: If an event $E$ can occur under mutually exclusive conditions $A_1, A_2, \ldots, A_k$, then the total probability of $E$ is $P(E) = \sum_{i=1}^{k} P(E|A_i)P(A_i)$. In this problem, the conditions are choosing Bag A or Bag B.
• Bayes' Theorem: Used to find the conditional probability of a "cause" ($A_i$) given an "effect" ($E$). The formula is $P(A_i|E) = \frac{P(E|A_i)P(A_i)}{P(E)}$. We are given $P(A|E)$ and need to use this to find an unknown variable.

2. Step-by-Step Solution

Step 1: Define Events and List Initial Probabilities & Bag Contents First, we clearly define the events and list all given information. This helps organize the problem.

• Let $A$ be the event that Bag A is chosen.
• Let $B$ be the event that Bag B is chosen.
• Let $E$ be the event that 2 balls drawn are 1 Red (R) and 1 Black (B).

Since one bag is chosen at random from two bags:

• $P(A) = \frac{1}{2}$
• $P(B) = \frac{1}{2}$

Now, let's detail the contents of each bag:

• Bag A: 2 White (W), 1 Black (B), 3 Red (R).
  • Total balls in Bag A = $2+1+3 = 6$ balls.
• Bag B: $n$ White (W), 3 Black (B), 2 Red (R).
  • Total balls in Bag B = $n+3+2 = n+5$ balls.

Step 2: Calculate Conditional Probabilities of Event E Next, we calculate the probability of drawing 1 Red and 1 Black ball, given which bag was chosen. This involves using combinations.

• Probability of drawing 1R and 1B given Bag A was chosen, $P(E|A)$: Bag A has 3 Red balls and 1 Black ball. We need to choose 1 Red from 3 and 1 Black from 1. The total number of ways to choose any 2 balls from 6 is $^{6}C_2$. $P(E|A) = \frac{{^{3}C_1} \times {^{1}C_1}}{{^{6}C_2}}$ We know $^{k}C_1 = k$ and $^{k}C_2 = \frac{k(k-1)}{2}$. $P(E|A) = \frac{3 \times 1}{\frac{6 \times 5}{2}} = \frac{3}{15} = \frac{1}{5}$

• Probability of drawing 1R and 1B given Bag B was chosen, $P(E|B)$: Bag B has 2 Red balls and 3 Black balls. We need to choose 1 Red from 2 and 1 Black from 3. The total number of ways to choose any 2 balls from $n+5$ is $^{n+5}C_2$. $P(E|B) = \frac{{^{2}C_1} \times {^{3}C_1}}{{^{n+5}C_2}}$ $P(E|B) = \frac{2 \times 3}{\frac{(n+5)(n+5-1)}{2}} = \frac{6}{\frac{(n+5)(n+4)}{2}} = \frac{12}{(n+5)(n+4)}$

Step 3: Calculate the Total Probability of Event E, $P(E)$ To apply Bayes' Theorem, we first need the overall probability of drawing 1 Red and 1 Black ball, $P(E)$, using the Total Probability Theorem. $P(E) = P(E|A)P(A) + P(E|B)P(B)$ Substitute the values from Step 1 and Step 2: $P(E) = \left(\frac{1}{5}\right) \times \left(\frac{1}{2}\right) + \left(\frac{12}{(n+5)(n+4)}\right) \times \left(\frac{1}{2}\right)$ $P(E) = \frac{1}{10} + \frac{6}{(n+5)(n+4)}$

Step 4: Apply Bayes' Theorem and Solve for $n$ We are given that the probability that both balls come from Bag A, given that they are 1 Red and 1 Black, is $\frac{6}{11}$. This is $P(A|E)$. Using Bayes' Theorem: $P(A|E) = \frac{P(E|A)P(A)}{P(E)}$ Substitute the given $P(A|E)$ and the expressions calculated in previous steps: $\frac{6}{11} = \frac{\left(\frac{1}{5}\right) \times \left(\frac{1}{2}\right)}{\frac{1}{10} + \frac{6}{(n+5)(n+4)}}$ $\frac{6}{11} = \frac{\frac{1}{10}}{\frac{1}{10} + \frac{6}{(n+5)(n+4)}}$

Now, let's solve this algebraic equation for $n$. To simplify, let $X = \frac{6}{(n+5)(n+4)}$. The equation becomes: $\frac{6}{11} = \frac{\frac{1}{10}}{\frac{1}{10} + X}$ Cross-multiply: $6 \left(\frac{1}{10} + X\right) = 11 \left(\frac{1}{10}\right)$ $\frac{6}{10} + 6X = \frac{11}{10}$ Subtract $\frac{6}{10}$ from both sides: $6X = \frac{11}{10} - \frac{6}{10}$ $6X = \frac{5}{10}$ $6X = \frac{1}{2}$ Divide by 6: $X = \frac{1}{12}$ Now, substitute back $X = \frac{6}{(n+5)(n+4)}$: $\frac{6}{(n+5)(n+4)} = \frac{1}{12}$ Cross-multiply: $(n+5)(n+4) = 6 \times 12$ $(n+5)(n+4) = 72$ Expand the left side to form a quadratic equation: $n^2 + 4n + 5n + 20 = 72$ $n^2 + 9n + 20 - 72 = 0$ $n^2 + 9n - 52 = 0$ Factor the quadratic equation. We need two numbers that multiply to -52 and add to 9. These numbers are 13 and -4. $(n+13)(n-4) = 0$ This gives two possible solutions for $n$: $n = -13$ or $n = 4$.

Since $n$ represents the number of white balls in Bag B, it must be a non-negative integer. Therefore, $n=-13$ is not a valid solution. Thus, $n=4$.

3. Common Mistakes & Tips

• Incorrect Application of Combinations: Ensure you correctly use $^k C_r$ for selecting balls, and remember to multiply combinations for independent selections (e.g., choosing 1 Red AND 1 Black).
• Algebraic Errors: Be meticulous with fraction arithmetic and solving the resulting equation. A small error can lead to an incorrect value of $n$.
• Ignoring Contextual Constraints: Always check if your final answer makes sense in the problem's context (e.g., $n$ must be a positive integer as it represents a count of balls).
• Confusing $P(A|E)$ with $P(E|A)$: Clearly understand which conditional probability is given and which needs to be calculated for Bayes' Theorem.

4. Summary

This problem is a comprehensive test of conditional probability, requiring the application of both the Total Probability Theorem and Bayes' Theorem. We systematically defined events, calculated probabilities of choosing each bag, determined the conditional probabilities of drawing the specified balls from each bag using combinations, and then used Bayes' Theorem with the given information to form an equation. Solving this quadratic equation led to the value of $n$. The logical derivation shows that $n=4$ is the consistent answer.

5. Final Answer

The final answer is $\boxed{4}$, which corresponds to option (C).